Lời giải:
Xét:
\(\frac{a^4}{(a+b)(a^2+b^2)}+\frac{b^4}{(b+c)(b^2+c^2)}+\frac{c^4}{(c+a)(c^2+a^2)}-\left[\frac{b^4}{(a+b)(a^2+b^2}+\frac{c^4}{(b+c)(b^+c^2)}+\frac{a^4}{(c+a)(c^2+a^2)}\right]\)
\(=\frac{a^4-b^4}{(a+b)(a^2+b^2)}+\frac{b^4-c^4}{(b+c)(b^2+c^2)}+\frac{c^4-a^4}{(c+a)(c^2+a^2)}=a-b+b-c+c-a=0\)
\(\Rightarrow \frac{a^4}{(a+b)(a^2+b^2)}+\frac{b^4}{(b+c)(b^2+c^2)}+\frac{c^4}{(c+a)(c^2+a^2)}=\frac{b^4}{(a+b)(a^2+b^2}+\frac{c^4}{(b+c)(b^+c^2)}+\frac{a^4}{(c+a)(c^2+a^2)}\)
\(\Rightarrow 2P=\frac{a^4+b^4}{(a+b)(a^2+b^2)}+\frac{b^4+c^4}{(b+c)(b^2+c^2)}+\frac{c^4+a^4}{(c+a)(c^2+a^2)}\)
Áp dụng hệ quả quen thuộc của BĐT AM-GM: \(x^2+y^2\geq \frac{(x+y)^2}{2}\) ta có:
\(a^4+b^4\geq \frac{(a^2+b^2)^2}{2}\)
\(a^2+b^2\geq \frac{(a+b)^2}{2}\)
\(\Rightarrow a^4+b^4\geq \frac{(a^2+b^2).\frac{(a+b)^2}{2}}{2}=\frac{(a^2+b^2)(a+b)^2}{4}\)
\(\Rightarrow \frac{a^4+b^4}{(a+b)(a^2+b^2)}\geq \frac{a+b}{4}\). Tương tự với các phân thức còn lại:
\(\Rightarrow 2P\geq \frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4}=\frac{a+b+c}{2}=2\)
\(\Rightarrow P\geq 1\). Vậy \(P_{\min}=1\Leftrightarrow a=b=c=\frac{4}{3}\)
\(a=b=c=\dfrac{4}{3}\Rightarrow P=1\)
Ta se cm \(P=1\) la GTNN cua P hay \(Σ\dfrac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge1\)
C-S: \(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{Σ\left(a+b\right)\left(a^2+b^2\right)}\)
Hay ta can cm bdt \(\dfrac{\left(a^2+b^2+c^2\right)^2}{Σ\left(a+b\right)\left(a^2+b^2\right)}\ge1=\dfrac{a+b+c}{4}\)
\(\Leftrightarrow4\left(a^2+b^2+c^2\right)^2\ge\left(a+b+c\right)\left(Σ\left(a+b\right)\left(a^2+b^2\right)\right)\)
\(\LeftrightarrowΣ\left(a-b\right)^2\left(a^2+b^2+c^2-ab\right)\ge0\)
